Find the sum of $6 + 11 + 16 +... + 4996 + 5001$.
Solution: Getting started We're dealing with an arithmetic series because the difference between terms is constant. That is, each term is $5$ greater than the one before it. We need a formula to compute the sum of the terms. Formula for arithmetic series The sum $S_n$ of a finite arithmetic series is $S_n = \dfrac {\left(a_1 + a_n \right)}{2} \cdot n$ where $a_1$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = 6)$ and the last term $(a_n = {5001})$ are given in the question. We need to find $n$ (the number of terms). Step 1: Find $n$ (the number of terms) The sequence increases by $5001 - 6 = 4995$ from the first term to the last term. Because the sequence increases by $5$ each time, it takes $\dfrac{4995}{5} = 999$ terms to get from the first term to the last term. We still need to count the first term, so there are $999 + 1 = {1000}$ terms in the sequence. In other words, $n = {1000}$. Step 2: Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac {\left(a_1 + a_n \right)}{2} \cdot n \\\\ S_{{1000}}&= \dfrac {\left(6 + {5001} \right)}{2} \cdot {1000} \\\\ S_{{1000}} &= 2503.5 \left(1000\right) \\\\ S_{{1000}} &= 2{,}503{,}500 \end{aligned}$ The answer $2{,}503{,}500$